/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 19 A resort area has three seafood ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 12: Problem 19

A resort area has three seafood restaurants, which employ students during the summer season. The local chamber of commerce took a random sample of five servers from each restaurant and recorded the tips they received on a recent Friday night. The results (in dollars) of the survey are shown in the table below. Assume that the Friday night for which the data were collected is typical of all Friday nights of the summer season. $$ \begin{array}{ccc} \hline \text { Barzini's } & \text { Hwang's } & \text { Jack's } \\ \hline 97 & 67 & 93 \\ 114 & 85 & 102 \\ 105 & 92 & 98 \\ 85 & 78 & 80 \\ 120 & 90 & 91 \\ \hline \end{array} $$ a. Would a student seeking a server's job at one of these three restaurants reject the null hypothesis that the mean tips on a Friday night are the same for all three restaurants? Use a \(5 \%\) level of significance. b. What will your decision be in part a if the probability of making a Type I error is zero? Explain.

Short Answer

Expert verified
a. The decision to reject the null hypothesis will depend on the calculated F-statistic and the critical value from the F-distribution table. b. With zero probability of making a Type I error, the decision will never be to reject the null hypothesis.

Step by step solution

01

Define the hypotheses

Start by defining the null hypothesis \(H_0\): the mean tips for all three restaurants are the same and the alternative hypothesis \(H_1\): the mean tips for at least one restaurant is different from the others.
02

Calculate the means and variances

Next, calculate the mean tip (\(x\)) for each restaurant and the overall mean tip (\(x_{overall}\)). Compute within-group variance (\(SSE\)) and between-group variance (\(SSB\)).
03

Calculate F-statistic and Critical value

Calculate the F-statistic using the formula: \(F = \frac{(SSB/(n-1))}{(SSE/(mn-n))}\), where m = number of groups (restaurants) and n=number of observations. Determine the critical value from F-distribution table with \((m-1, mn-m)\) degrees of freedom at \(5\%\) significance level.
04

Make decision

Reject the null hypothesis if \(F > Critical Value\). This means that if the calculated F statistic is greater than the critical value obtained from the F-distribution table, there is significant evidence to reject the null hypothesis.
05

Decision for Zero Probability of Type I error

For zero probability of Type I error, we would never reject the null hypothesis. This is because a Type I error is made when the null hypothesis is true but is rejected, so ensuring zero probability of a Type I error means never rejecting the null hypothesis regardless of the results we obtain.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Hypothesis Testing
Hypothesis testing is a fundamental concept in statistics for making inferences about a population based on sample data. It consists of two key elements: the null hypothesis ( H_0 ) and the alternative hypothesis ( H_1 ). The null hypothesis usually states that there is no effect or difference, and in our exercise, it signifies that the mean tips for all three restaurants are the same. The alternative hypothesis suggests there is a difference, indicating that at least one restaurant's mean tip is different from the others.

In hypothesis testing, we collect data and determine if there is enough statistical evidence to reject the null hypothesis. The process involves choosing a significance level, analyzing the data, calculating relevant test statistics, and comparing them to critical values derived from statistical distributions. If the test statistic falls within the critical region, we reject the null hypothesis, concluding there is a significant difference.
F-statistic
The F-statistic is a ratio that compares two variances to determine if they are significantly different. In the context of ANOVA (Analysis of Variance), the F-statistic is used to test whether different samples have the same variance, which implies they could have the same mean. The ANOVA uses the F-statistic to compare between-group variance (the variance between the means of the groups) to the within-group variance (the variance within each group).
  • Between-group variance (SSB) accounts for variations due to interactions between the groups.
  • Within-group variance (SSE) reflects the variance within each group.
The F-statistic formula is: \(F = \frac{(SSB/(m-1))}{(SSE/(mn-n))}\), where \(m\) is the number of groups (restaurants) and \(n\) is the total number of observations. A larger F-statistic indicates greater variation between group means than within groups, suggesting significant differences that might lead to rejection of the null hypothesis.
Type I Error
A Type I error occurs when the null hypothesis is wrongly rejected when it is actually true. This mistake is an integral part of hypothesis testing, as it represents a false positive result.
  • In our exercise, a Type I error would mean concluding that not all restaurants have the same mean tips, although they do.
  • It's crucial to control the probability of making a Type I error, which is directly related to the significance level chosen.
Choosing a significance level helps to set the threshold for when we'll reject the null hypothesis, effectively balancing the risk of making a Type I error. But if a zero probability of Type I error is required, the null hypothesis should never be rejected, meaning no conclusion of difference should ever be made without absolute evidence.
Significance Level
The significance level, denoted by \(\alpha\), is a threshold set before conducting a hypothesis test. It represents the probability of making a Type I error, or falsely rejecting the null hypothesis when it is true. Commonly used significance levels are \(5\%\), \(1\%\), or \(10\%\). In our exercise, a \(5\%\) significance level is used.
  • This means there is a \(5\%\) risk of rejecting a true null hypothesis.
  • The choice of significance level affects the critical value and thus the decision rule regarding the null hypothesis.
A lower significance level means stricter criteria for rejecting the null hypothesis, reducing the risk of a Type I error but potentially increasing risk of a Type II error (failing to reject a false null hypothesis). Therefore, the choice of significance level is a balancing act in statistical testing.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Describe the assumptions that must hold true to apply the oneway analysis of variance procedure to test hypotheses.

A large company buys thousands of lightbulbs every year. The company is currently considering four brands of lightbulbs to choose from. Before the company decides which lightbulbs to buy, it wants to investigate if the mean lifetimes of the four types of lightbulbs are the same. The company's research department randomly selected a few bulbs of each type and tested them. The following table lists the number of hours (in thousands) that each of the bulbs in each brand lasted before being burned out. $$ \begin{array}{cccc} \hline \text { Brand I } & \text { Brand II } & \text { Brand III } & \text { Brand IV } \\ \hline 23 & 19 & 23 & 26 \\ 24 & 23 & 27 & 24 \\ 19 & 18 & 25 & 21 \\ 26 & 24 & 26 & 29 \\ 22 & 20 & 23 & 28 \\ 23 & 22 & 21 & 24 \\ 25 & 19 & 27 & 28 \\ \hline \end{array} $$ At a \(2.5 \%\) significance level, test the null hypothesis that the mean lifetime of bulbs for each of these four brands is the same.

Determine the critical value of \(F\) for the following. a. \(d f=(6,12)\) and area in the right tail \(=.01\) b. \(d f=(6,40)\) and area in the right tail \(=.01\) c. \(d f=(6,100)\) and area in the right tail \(=.01\)

A consumer agency wanted to find out if the mean time taken by each of three brands of medicines to provide relief from a headache is the same. The first drug was administered to six randomly selected patients, the second to four randomly selected patients, and the third to five randomly selected patients. The following table gives the time (in minutes) taken by each patient to get relief from a headache after taking the medicine. $$ \begin{array}{ccc} \hline \text { Drug I } & \text { Drug II } & \text { Drug III } \\ \hline 25 & 15 & 44 \\ 38 & 21 & 39 \\ 42 & 19 & 54 \\ 65 & 25 & 58 \\ 47 & & 73 \\ 52 & & \\ \hline \end{array} $$ At a \(2.5 \%\) significance level, will you reject the null hypothesis that the mean time taken to provide relief from a headache is the same for each of the three drugs?

An ophthalmologist is interested in determining whether a golfer's type of vision (far-sightedness, near-sightedness, no prescription) impacts how well he or she can judge distance. Random samples of golfers from these three groups (far-sightedness, near-sightedness, no prescription) were selected, and these golfers were blindfolded and taken to the same location on a golf course. Then each of them was asked to estimate the distance from this location to the pin at the end of the hole. The data (in yards) given in the following table represent how far off the estimates (let us call these errors) of these golfers were from the actual distance. A negative value implies that the person underestimated the distance, and a positive value implies that a person overestimated the distance. $$ \begin{array}{l|rrrrrrrrr} \hline \text { Far-sighted } & -11 & -9 & -8 & -10 & -3 & -11 & -8 & 1 & -4 \\\ \hline \text { Near-sighted } & -2 & -5 & -7 & -8 & -6 & -9 & 2 & -10 & -10 \\\ \text { No prescription } & -5 & 1 & 0 & 4 & 3 & -2 & 0 & -8 & \\ \hline \end{array} $$ At a \(1 \%\) significance level can you reject the null hypothesis that the average errors in predicting distance for all golfers of the three different vision types are the same.
See all solutions

Recommended explanations on Math Textbooks

Pure Maths

Read Explanation

Logic and Functions

Read Explanation

Decision Maths

Read Explanation

Probability and Statistics

Read Explanation

Mechanics Maths

Read Explanation

Calculus

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.